Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $p = \dfrac{10k - 30}{3k - 24} \times \dfrac{k^2 - 5k - 24}{k^2 - 3k} $
Solution: First factor the quadratic. $p = \dfrac{10k - 30}{3k - 24} \times \dfrac{(k - 8)(k + 3)}{k^2 - 3k} $ Then factor out any other terms. $p = \dfrac{10(k - 3)}{3(k - 8)} \times \dfrac{(k - 8)(k + 3)}{k(k - 3)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ 10(k - 3) \times (k - 8)(k + 3) } { 3(k - 8) \times k(k - 3) } $ $p = \dfrac{ 10(k - 3)(k - 8)(k + 3)}{ 3k(k - 8)(k - 3)} $ Notice that $(k - 3)$ and $(k - 8)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ 10\cancel{(k - 3)}(k - 8)(k + 3)}{ 3k\cancel{(k - 8)}(k - 3)} $ We are dividing by $k - 8$ , so $k - 8 \neq 0$ Therefore, $k \neq 8$ $p = \dfrac{ 10\cancel{(k - 3)}\cancel{(k - 8)}(k + 3)}{ 3k\cancel{(k - 8)}\cancel{(k - 3)}} $ We are dividing by $k - 3$ , so $k - 3 \neq 0$ Therefore, $k \neq 3$ $p = \dfrac{10(k + 3)}{3k} ; \space k \neq 8 ; \space k \neq 3 $